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As the title says, I want to show that the limit of

$$\lim_{x\to 0} \frac{\cos(x)}{x}$$

doesn't exist.

Now for that I'd like to show in a formally correct way that

$$\lim_{x\to 0^+} \frac{\cos(x)}{x} = +\infty$$

I'm sure this is right since $\displaystyle\lim_{x\to 0^+} \cos(x) = 1$ and $\displaystyle\lim_{x\to 0^+} x = 0$, but since $\displaystyle\lim_{x\to 0^+} x = 0$ I can't just say:

$$\lim_{x\to 0^+} \frac{\cos(x)}{x} = \frac{1}{\displaystyle\lim_{x\to 0^+} x}$$

Things I tried: expand the fraction and use L'Hospital's rule (This didn't seem to yield results even though I tried quite a few things, should it? If it does, should it always work? It worked for similar problems) and going by the definition of the limit which didn't work for me either.

I'm grateful for any hints for doing this problem and similar problems especially.

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    $\begingroup$ I would pick an arbitrary large number $N$, and show that whenever $x$ is less than a threshold value, $\cos(x)/x > N$. $\endgroup$ – Patrick Li Jun 9 '13 at 15:35
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    $\begingroup$ L'Hôpital's rule fails here because its assumptions aren't satisfied. Briefly, it works for $\infty/\infty$ or $0/0$ type limits, and this one is of the kind $1/0$. This is much simpler than L'Hôpital, in fact. $\endgroup$ – Harald Hanche-Olsen Jun 9 '13 at 16:23
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Hint: You can find a region near $x=0$ where $\cos x \gt \frac 12$

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If $|x| \in [0,\frac{\pi}{4}]$, then $\cos x \ge \frac{1}{2}$. Then for $n>1$ we have $\frac{1}{n} \in [0,\frac{\pi}{4}]$, hence setting $x_n = \frac{1}{n}$ we have $\frac{\cos x_n}{x_n} \ge \frac{1}{2 x_n} = \frac{n}{2}$.

Similarly, with $x'_n = -\frac{1}{n}$, $\frac{\cos x'_n}{x'_n} \le \frac{1}{2 x'_n} = -\frac{n}{2}$.

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